*****************************
.5k*y^2 = .5k*x^2 + m*g*(x+a/2)
This is the "true" elastic PE PLUS the gravitational PE
measured from x= -a/2 !
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----- Original Message -----
Sent: Tuesday, December 04, 2001 9:52
PM
Subject: Re: test problem
To perhaps clarify further aspects of this type
problem:
As before, let x = the mass position as measured (up is
positive) from the relaxed spring position.
Let x=-a (a will be positive) be the equilibrium position
for the hanging mass, so that ka=mg.
Let y = the mass position as measured (up is positive) from
this (x=-a) equilibrium position, so that y = x + a.
Then N2 gives for the DE of motion:
mx'' = -k*x - mg ; Now substitute x = y-a
=>
my'' = -k*y + k*a -mg ; Now use ka = mg
=>
my'' = -k*y => SHM about the point y=0
(x=-a)
Note also that the "pseudo" elastic PE of this y oscillation
is:
.5 K*y^2 = .5 k*x^2 + .5*k*a^2 + kax ; Now use ka=mg
=>
,5k*y^2 = .5k*x^2 +.5 m*g*(x+a)
This is the "true" elastic PE PLUS the gravitational PE
measured from x= -a !
IE: You can use the "pseudo" elastic PE as the TOTAL
system PE (Cuz it is so, numerically, if the arbitrary zero of
gravitational PE is properly chosen).